3條f.1數學方程

1.

7k-(9k-16) = 14-(19-k)
7k-9k+16 = 14-19+k
-3k = -21
k = 7




2.

2[p-2(p+1)] = 8p+25
2(p-2p-2)-8p = 25
2p-4p-4-8p = 25
-10p = 29
p = -2.9




3.

3[4(2.5-q)+1] = 13-2q
12(2.5-q)+3 = 13-2q
30-12q+3-13+2q = 0
10q = 20
q = 2

1. 7k-(9k-16)=14-(19-k)
7k-9k+16=14-19+k
7k-9k+k=14-19-16
-k=-21
k=21

2 2[p-2(p+1)]=8p+25
2(p-2p-2)=8p+25
2p-4p-4=8p+25
-2p-8p=4+25
-10p=29
-p=2.9
p=-2.9

3. 3[4(2.5-q)+1]=13-2q
3(10-4q+1)=13-2q
30-12q+3=13-2q
-10q=-20
q=2

希望冇計錯啦!
其實只要將d正負號搞清楚,
細括號個d乘先,
仲有如果係方程既話,
調過隔離要將負變正,正變負,
記住"負負得正,正正得正,正負得負"
就唔會有錯架啦!

1. 7k-(9k-16)=14-(19-k)

2. 2[p-2(p+1)]=8p+25

3. 3[4(2.5-q)+1]=13-2q

1. 7k-(9k-16)=14-(19-k)
7k-9k-16=14-19-k
7k-9k+k=14-19+16
-k=11

2.2[p-2(p+1)]=8p+25
2(p-2p+2)=8p+25
2p-4p+2=8p+25
2p-4p-8p=25-2
-10p=23
p=-23/10

3. 3[4(2.5-q)+1]=13-2q
3(10-4q+1)=13-2q
30-12q+3=13-2q
-12q+2q=13-3-30
-10q=-20
q=-20/-10
q=2
hope i can help you!